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Topic: getting rid of H's  (Read 5947 times)

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Offline Seks

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getting rid of H's
« on: April 13, 2008, 03:15:25 PM »
H2 in Pt to alkene will make it alkane (hydrogenation)

How do I go about making an alkane into an alkene by getting rid of 2 adjacent H?

Offline sjb

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Re: getting rid of H's
« Reply #1 on: April 13, 2008, 03:24:06 PM »
As far as I know there is no general way of doing this.

But in certain circumstances you may be able to

a) free radical brominate then dehydohalogenate with base.

b) conversion of aldehydes or ketones to enals (and enones) via selenate oxidation and elimination.

c) HVZ halogenation of acids then elimination.

d) b) free radical (?) hydroxylate then dehydrate

Do you have a particular alkane -> olefin conversion in mind?

Offline english

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Re: getting rid of H's
« Reply #2 on: April 13, 2008, 03:25:58 PM »
I believe the Wittig reaction (pronounced "Vit-tig") is one of many methods of producing an alkene.  Dehydrogenation is another, which I think may be used on the industrial scale.

Offline Seks

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Re: getting rid of H's
« Reply #3 on: April 13, 2008, 03:27:29 PM »
well the practice question is converting a cyclopentane to ethoxycyclopentane

I came up with:

1.  Br2 in light to give Bromocyclopentane
2.  H2O of product #1 to give cyclopentanol
3.  NaH then CH3CH2Br of product #2 to give the target compound

 ???

Offline Arkcon

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Re: getting rid of H's
« Reply #4 on: April 13, 2008, 03:27:44 PM »
Ah, you want a dehydrogenation, now that is a tricky one, usually you start with a feedstock that already has double or triple bonds.  There are some specialized catalysts for this purpose, as I understand it some rare-earth oxides, at high temps, will steal two hydrogens from a carbon chain, producing a double bond and water.

*[EDIT]*
Oh, the gang's all here, aren't they?  We should just use the shout out function.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline english

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Re: getting rid of H's
« Reply #5 on: April 13, 2008, 03:49:45 PM »
3.  NaH then CH3CH2Br of product #2 to give the target compound
The hydride (NaH) would just deprotonate the alcohol.  A deprotonated alcohol is a very poor nucleophile, so I wouldn't consider it performing a substitution.

Offline agrobert

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Re: getting rid of H's
« Reply #6 on: April 13, 2008, 04:20:52 PM »
well the practice question is converting a cyclopentane to ethoxycyclopentane

I came up with:

1.  Br2 in light to give Bromocyclopentane
2.  H2O of product #1 to give cyclopentanol
3.  NaH then CH3CH2Br of product #2 to give the target compound

 ???

This will work but it has nothing to do with original post.  Bromination by radical chemistry, hydrolysis and then Williamson ether all sounds good.  You may have to push the ether synthesis because your alcohol is 2ยบ.

3.  NaH then CH3CH2Br of product #2 to give the target compound
The hydride (NaH) would just deprotonate the alcohol.  A deprotonated alcohol is a very poor nucleophile, so I wouldn't consider it performing a substitution.

Are you sure?
http://en.wikipedia.org/wiki/Williamson_ether_synthesis
In the realm of scientific observation, luck is only granted to those who are prepared. -Louis Pasteur

Offline english

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Re: getting rid of H's
« Reply #7 on: April 13, 2008, 04:28:58 PM »
Are you sure?
http://en.wikipedia.org/wiki/Williamson_ether_synthesis
Oh lord...how easily we forget.   ::)

Ha ha, I see what I did...for some reason I was too focused on the actual hydroxyl group and not the fact it's a alkoxide, which are fairly decent nucleophiles.

That's what too much NMR will do to you I guess.  ;)
« Last Edit: April 13, 2008, 04:41:20 PM by g english »

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